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If a three-digit positive integer has its digits reversed, the resulting three-digit positive integer is less than the original integer by 297. How many such pairs are possible?
A. 3
B. 6
C. 7
D. 60
E. 70
M37-46
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A. 3
B. 6
C. 7
D. 60
E. 70
M37-46
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Official Solution:
If a three-digit positive integer has its digits reversed, the resulting three-digit positive integer is less than the original integer by 297. How many such pairs are possible?
A. \(3\)
B. \(6\)
C. \(7\)
D. \(60\)
E. \(70\)
Say the original three-digit number is \(abc\) and the reversed number is \(cba\).
Given: \(abc-cba=297\);
\((100a + 10b + c ) - (100c + 10b + c) = 297\);
\(a-c=3\).
\((a, c)\) can be (9, 6), (8, 5), (7, 4), (6, 3), (5, 2), and (4, 1) (6 pairs). Notice that (6, 0) pair is not possible: \(c\) cannot be 0 because in this case \(cba\) would be a two-digit number, not a three-digit number as given in the stem.
Finally, don't forget about \(b\). It can take 10 values: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
So, there are a total of \(6*10=60\) numbers are possible.
Answer: D
_________________
If a three-digit positive integer has its digits reversed, the resulting three-digit positive integer is less than the original integer by 297. How many such pairs are possible?
A. \(3\)
B. \(6\)
C. \(7\)
D. \(60\)
E. \(70\)
Say the original three-digit number is \(abc\) and the reversed number is \(cba\).
Given: \(abc-cba=297\);
\((100a + 10b + c ) - (100c + 10b + c) = 297\);
\(a-c=3\).
\((a, c)\) can be (9, 6), (8, 5), (7, 4), (6, 3), (5, 2), and (4, 1) (6 pairs). Notice that (6, 0) pair is not possible: \(c\) cannot be 0 because in this case \(cba\) would be a two-digit number, not a three-digit number as given in the stem.
Finally, don't forget about \(b\). It can take 10 values: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
So, there are a total of \(6*10=60\) numbers are possible.
Answer: D
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General Discussion
sahilvijay
Joined: 29 Jun 2017
Last visit: 15 Apr 2021
Posts: 288
GPA: 4
WE:Engineering (Transportation)
Products:
Posts: 288
Updated on: 17 Dec 2017, 02:49
Originally posted by sahilvijay on 07 Sep 2017, 04:17.
Last edited by sahilvijay on 17 Dec 2017, 02:49, edited 2 times in total.
Last edited by sahilvijay on 17 Dec 2017, 02:49, edited 2 times in total.
IMO D
abc -cba =297
100a+10b+c - (100c+10b+a) = 297
100(a-c) +c-a = 297
99(a-c) =297
a-c =3
b can take 0-9 => 10 values
but c can not be 0 since reverse number is also 3 digit number
c max = 6 because 6+3 =9 =a max
so c can take 1,2,3,4,5,6 => 6 values
Hence total numbers possible
6x10=60
D is the Answer
abc -cba =297
100a+10b+c - (100c+10b+a) = 297
100(a-c) +c-a = 297
99(a-c) =297
a-c =3
b can take 0-9 => 10 values
but c can not be 0 since reverse number is also 3 digit number
c max = 6 because 6+3 =9 =a max
so c can take 1,2,3,4,5,6 => 6 values
Hence total numbers possible
6x10=60
D is the Answer
Bunuel
100x+10y+z-100z+10y+x=297
99(x-z)=297
x-z = 3
there are 6 possible numbers for x. for example: 9y6-6y9=297 and so on till x=4
as for y. there are 10 possibilities: from 0 to 9
so 6*10=60 pairs possible.
so answer is D
Ans: D
N=100x+10y+z
=> (100x+10y+z) - (100x+10y+z) =297
=> x-z =3
(1) x <=9, z <=6 <=> Total 6 pairs z=1,2,3,4,5 & 6
(2) y can be 0 to 9(Total 10)
Total possibilities = 6*10=60
N=100x+10y+z
=> (100x+10y+z) - (100x+10y+z) =297
=> x-z =3
(1) x <=9, z <=6 <=> Total 6 pairs z=1,2,3,4,5 & 6
(2) y can be 0 to 9(Total 10)
Total possibilities = 6*10=60
sahilvijay
What about a?
b*c= 10 * 6= 60
so what about a? where have it gone?
sahilvijay
Joined: 29 Jun 2017
Last visit: 15 Apr 2021
Posts: 288
GPA: 4
WE:Engineering (Transportation)
Products:
Posts: 288
lor12345
No need to consider a - above solution is self explanatory
Answer is D
Possible combinations are 4b1, 5b2, 6b3,7b4,8b5, 9b6
B can be any dogit from 0-9 , hence 10 possiblities. and the above combination are only 6 , therefore answer is 6*10 = 60
Possible combinations are 4b1, 5b2, 6b3,7b4,8b5, 9b6
B can be any dogit from 0-9 , hence 10 possiblities. and the above combination are only 6 , therefore answer is 6*10 = 60
Bunuel
xyz-zyx=297
297/99=3=x-z
6 possible x-z combinations: 9-6,8-5,7-4,6-3,5-2,4-1
10 possible y values: 0-9
6*10=60 possible pairs
D
Bunuel
Par of GMAT CLUB'S New Year's Quantitative Challenge Set
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Kinshook
Major Poster
Joined: 03 Jun 2019
Last visit: 04 Sep 2026
Posts: 6,129
Location: India
Schools: HBS '26 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
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Bunuel
Asked: If a three-digit positive integer has its digits reversed, the resulting three-digit positive integer is less than the original integer by 297. How many such pairs are possible?
Let the 3-digit positive integer be of the form xyz
(100x + 10y + z) - (100z + 10y + x) = 99(x-z) = 297
x -z = 3
Since x & z are not 0
(x,z) = {(4,1),(5,2),(6,3),(7,4),(8,5),(9,6)} = 6 cases
y can take 10 values
Total such numbers = 6*10 = 60
IMO D
Bunuel
\( ABC-CBA=297:100a+10b+c-100c-10b-a=297, 99(a-c)=297, (a-c)=3\)
\((A,C)=(9,6;8,5;7,4;6,3;5,2;4,1)…(A,C)≠(3,0):C>0\)
\((A,C)=6.pairs…B=(0,1,2,3,4,5,6,7,8,9)=10\)
\(Total.cases:6*10=60\)
Ans (D)
Considering the three-digit number to be ABC.
The value of the three-digit number is given by 100*A+10*B+C.
When the digits are reversed the number becomes CBA.
The value becomes : 100*C + 10*B+A.
Since the value decreased by 297 this can be represented as :
\(\left(100A+10B+C\right)-\left(100C+10B+A\right)\ =\ 297\)
99(A-C) = 297.
A-C = 3.
The different possibilities include : (A = 9, C = 6), (A = 8, C = 5), (A = 7, C = 4), (A = 6, C = 3), (A = 5, C = 2), (A = 4, C = 1)
A and C cannot take the value of 0 because then the numbers do not satisfy the three digit number condition.
For the 6 case B can take any value from 0 to 9.
A total of 10*6 = 60 cases.
The value of the three-digit number is given by 100*A+10*B+C.
When the digits are reversed the number becomes CBA.
The value becomes : 100*C + 10*B+A.
Since the value decreased by 297 this can be represented as :
\(\left(100A+10B+C\right)-\left(100C+10B+A\right)\ =\ 297\)
99(A-C) = 297.
A-C = 3.
The different possibilities include : (A = 9, C = 6), (A = 8, C = 5), (A = 7, C = 4), (A = 6, C = 3), (A = 5, C = 2), (A = 4, C = 1)
A and C cannot take the value of 0 because then the numbers do not satisfy the three digit number condition.
For the 6 case B can take any value from 0 to 9.
A total of 10*6 = 60 cases.
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